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CHEMISTRY
IUPAC NOMENCLATUREAND ISOMERISM
IUPAC Nomenclature
1. HISTORICAL INTRODUCTIONIt was observed that initially compounds were found from two sources living organism and non living matter.
The compounds which were derived from living organism were called as organic compounds while those from
non living matter were named as inorganic compounds. Berzilius thought that organic compounds were
produced from their element by laws different from those governing the formation of inorganic compounds.
This led him to the belief that organic compounds are formed under the influence of a vital force and they
couldn't be prepared artificially.This myth continued till Wohler converted inorganic compound ammonium
cyanate found from non living matters into urea an organic compound.
NH4CNO heat
NH2CONH
2
ammonium cyanate urea
The myth of vital force theory was completely ended with the synthesis of acetic acid from its elements by
Kolbe in 1845,and the synthesis of methane by Berthelot in 1856.
Since then millions of compounds have been produced in laboratory. Basically organic compounds are made
of essentially carbon and other elements like Hydrogen, Oxygen, Nitrogen Phosphorus, Sulphur etc..
2. SOME IMPORTANT DEFINITIONS
(i) Catenation : The property of atoms of an element to link with one another forming chains of identical
atoms is called catenation.
(ii) Homologous series : Homologous series may be defined as a series of similarly constituted compounds
in which the members possess the same functional group, have similar chemical characteristics and have a
regular gradation in their physical properties. The two consecutive members differ in their molecular formula
by CH2.
(iii) Isomerism : Compounds which have the same molecular formula but differ in physical and chemical
properties are called isomers and the phenomena is called isomerism.
Isomerism can be broadly classified into two categories
1. Structural Isomerism
2. Stereoisomerism
3. CLASSIFICATION OF ORGANIC COMPOUNDSIf we look at the vast number of organic compounds, two things will strike our mind –the skeleton is either open
or closed. Based on this, organic compounds are classified as acyclic (open–chain) or cyclic (closed–chain)
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e.g.
1. Alkanes CH3– CH
2– CH
3Propane
2. Alkenes CH3– CH = CH
2Propene
3. Alkynes HC CH Ethyne
4. Aliphatic compounds with other functional groups CH3– CH
2– OH Ethanol
CH3– CH
2– COOH Propanoic acid
5. Homo alicyclic compound Cyclohexane
6. Heteroalicyclic compoundO
Pentahydropyran
7. Heterocyclic aromatic compound Pyridine
8. Homo cyclic benzenoid aromatic compounds
OH
Phenol
Napthalene
9. Homo cyclic Non-benzenoid aromatic compound [18] annulene
4. ORGANIC COMPOUNDS AND FUNCTIONAL GROUPSNumber of known organic compounds is much more than inorganic compounds but, it has been possible to
group them into classes or families based on their structural features. This has given organic chemistry a
logical and systematic shape.
In these compounds, there are a number of atoms or groups which show constancy of properties. Such
groups are called functional groups. Examples of functional groups are :
4.1 AlkanesThese are open–chain aliphatic compounds which have no functional groups. These are
represented by the general formula, CnH
2n + 2where n = 1, 2, 3, ............
The names of the alkanes for n = 1, 2, ..........., 10 must be committed to memory since all of organic
nomenclature is related directly or indirectly to the name of alkane CnH
2n + 2
n = 1 means CH4
– Methane
n = 2 means C2H
6– Ethane
n = 3 means C3H
8– Propane
n = 4 means C4H
10– Butane
n = 5 means C5H
12– Pentane
n = 6 means C6H
14– Hexane
n = 7 means C7H
16– Heptane
n = 8 means C8H
18– Octane
n = 9 means C9H
20– Nonane
n = 10 means C10
H22
– Decane
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4.2 Alkyl GroupsWhen we remove a hydrogen from alkane, we get an alkyl group :
General formula 1n2nH–
2n2n HCHC
alkylalkaneyl
ane–
e.g. methylmethaneyl
ane–
3H–
4 CHCH
e.g. propylpropaneyl
ane–
73H–
83 HCHC
4.3 AlkenesAlkenes are open chain hydrocarbons and having carbon–carbon double bonds, C = C. They have the
general formula CnH
2nwhere n = 2, 3, 4 ... etc. and are also called alkylenes or olefins. The first three
members are commonly named by their common names
e.g.
CH2= CH
2CH
3– CH = CH
2CH
3– CH
2– CH = CH
2 23
3
CHCCH|CH
ethylene propylene butylene Isobutylene
4.4 Alkenyl GroupsLike alkyl groups (alkane – H), there are three commonly encountered alkenyl groups which are given
common names.
e.g.
CH2
= CH2
Hremove
CH2
= CH –
vinyl group
4.5 AlkynesUnsaturated aliphatic hydrocarbons containing a carbon–carbon triple bond are called alkynes. In the common
system, they are called acetylenes, after the name of the first member of this family. i.e. acetylene
General formula CnH
2n-2
where n = 2, 3, 4....... etc.
Common Names Acetylene and its alkyl derivatives.
IUPAC names
Alkane – ane + yne = Alkyne
The position of the triple bond on the parent chain is designated by lowest possible arabic numerals
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The common IUPAC names of a few simple alkynes are given below.
n = 2 CH CH Acetylene
n = 3 CH3–CCH Methyl acetylene
n = 4 CH3– CH
2– C CH Ethylacetylene
n = 6 CH – C C – CH3
CH3
CH3
Methyl isopropyl acetylene
4.6 Alkynyl Groups
HC CH H HC C –
(-e + yl = Ethynyl)
H3C – C CH H H
3C – C C –
(-e + yl = propynyl)
4.7 Functional group and ResidueThe characteristic group of atom which decided the physical and chemical properties of an organic molecule
is called functional group.
Functional group is that portion of molecule which is highly reactive and takes part in chemical reactions.
Rest of the molecule is called Residue.
e.g. CH3CH
2CH
2CH
2– COOH
Residue FunctionalGroup
4.8 Functional groups
Class of functional groups Name
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
R – COOH
R – SO H3
R – C – O – C – R
R – COOR
R – C – X
R – C – NH2
R – C N
R – C – H
R – C – R
R – OH
R – SH
R – NH2
Sulfonic acid
Anhydride
Ester
Acid halide
Amide
Alkane nitrile
Aldehyde
Ketone
Alcohol
Thiol
Amine
|| ||
||
||
||
||
O O
O
O
O
O
Carboxylic acid
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Examples of Compound containing different functional groups
(1) – COOH group
(i) HCOOH Formic acid
(ii) CH3COOH Acetic acid
(iii) CH3– CH
2– COOH Propionic acid
(2) – SO3H group
(i) CH3SO
3H Methane sulfonic acid
(ii) CH3CH
2SO
3H Ethane sulfonic acid
(iii) CH3– CH
2– CH
2– SO
3H Propane sulfonic acid
(3)
OO||||COC group
(i) 33 CHCOCCH||||OO
Acetic anhydride
(ii) 3223 CHCHCOCCHCH||||OO
Propionic anhydride
(iii)322223 CHCHCHCOCCHCHCH
||||OO
Butanoic anhydride
(4)
O||
ROC group
(i)3OCHCH
||O
Methyl formate
(ii)52HCOCH
||O
Ethyl formate
(iii)523 HCOCCH
||O
Ethyl acetate
(5)
O||
XC group
(i)
O||
ClCH Formyl chloride
(ii)
O||
BrCCH3 Acetyl bromide
(iii)
O||
ClCCHCHCH 223 n-Butyryl chloride
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(6)
O||
NHC 2 group
(i)2NHCH
||O
Formamide
(ii) 23 NHCCH||O
Acetamide
(iii)223 NHCCHCH
||O
Propanamide
(7) – C N group
(i) CH3– CN Methyl cyanide or Acetonitrile
(ii) CH3– CH
2– CN Ethyl cyanide or propionitrile
(iii) CH3– CH
2– CH
2– CN n–Propyl cyanide or n–Butyronitrile
(8)
O||
HC group
(i) HCH||O
Formaldehyde
(ii) CH3– CHO Acetaldehyde
(iii) CH3– CH
2– CH
2– CHO n-Butyraldehyde
(9)
O||
RC group
(i)
O||
CHCCH 33 Dimethyl ketone or acetone
(ii)
O||
CHCHCCH 323 Ethylmethyl ketone
(iii)
O||
CHCHCHCCH 3223 Methyl n–propyl ketone
(10) – OH group
(i) CH3– OH Methyl alcohol
(ii)
OH|
CHCHCH 33 Isopropyl alcohol
(iii) CH3– CH
2– CH
2– OH n-Propylalcohol
(11) – SH group
(i) CH3– SH Methane thiol
(ii) CH3– CH
2– SH Ethane thiol
(iii) CH3– CH
2– CH
2– SH Propane thiol
(12) – NH2group
(i) CH3– NH
2Methylamine or Aminomethane
(ii) CH3– CH
2– NH – CH
3Ethyl methyl amine or N-Methylaminoethane
(iii) (CH3)3N Trimethylamine or N, N–Dimethyl aminomethane
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5. SOME COMMON NAMES OF HYDROCARBON ALKYL GROUPS5.1 Iso Alkyl group
A compound having3
3
CHCH|CH
group is called iso alkyl group
e.g.3
3
CH|CHCH
3
23
CH|
CHCHCH
3
323
CH|
CHCHCHCH
Iso propyl Iso butyl Iso pentane
Note: Iso alkyl group name can be used in IUPAC Nomenclature and its first letter 'I' is consider for alphabetical
seniority
e.g.
3
33
2
432
3
1
CH|
CHCHCH||
HCHCHCHC
3
6
2
5
HCHC
3–Isopropyl–2–methylhexane
5.2 Neo Alkyl group
Compound having
3
23
3
CH|
CHCCH|
CH
group is called neo alkyl group.
e.g.
3
23
3
CH|
–CHCCH|CH
Neopentyl
3
33
3
CH|
CHCCH|CH
Neopentane
3
323
3
CH|
CHCHCCH|
CH
Neohexane
6. TYPES OF CARBON AND HYDROGEN ATOMS IN ALKANES
The carbon atoms in an alkane molecule may be classified into four types as primary (1°), secondary (2°),
tertiary (3°) and quaternary (4°) as follows
(i) A carbon atom attached to one another (or no other) carbon atom is called a primary carbon atom and is
designated as 1° carbon
e.g. (i) 3
1
3
1
HCHC
(ii) 3
1
23
1
HCCHHC
(ii) A carbon atom attached to two other carbon atoms is called a secondary carbon atom and is designated as
2° carbon
e.g. (i)32
2
3 HCHCHC
(ii) 32
2
2
2
3 CH–HC–HC–CH
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(iii) A carbon atom attached to three other carbon atoms is called a tertiary carbon atom and is designated as 3°
carbon.
e.g. (i) 33
3
3
CHHCCH|CH
(ii) 323
3
3
CHCHHCCH|CH
(iv) A carbon atom attached to four other carbon atoms is called a quaternary carbon atom and is designated as
4° carbon
e.g. (i)
3
34
3
3
CH|
CHCCH
|CH
(ii)
32
324
3
3
CHCH|
CHCHCCH
|CH
The hydrogen atoms attached to 1°, 2° and 3° carbon atoms are called primary (1°) secondary (2°) and
tertiary (3°) hydrogen atoms. It may be noted here that there is nothing like quaternary hydrogen atom. Since
a quarternary carbon does not carry any hydrogen.
Ex.1 How many 1°, 2°, 3° and 4° carbon atoms are present in following molecule.
3233
323
3
CH–CHCHCH|||
CH–CH–CH–C–CH–CH|
CH
Sol. 1° Carbon atoms = 6 , 2° Carbon atoms = 2 , 3° Carbon atoms = 2 , 4° Carbon atom = 1
Note : Primary, secondary, tertiary & quaternary carbon atoms in a molecule are denoted by the letters p, s, t and
q respectively.
Ex.2 How many 1°, 2°, 3° and 4° carbon atoms are present in following molecule
33
323
3
CHCH||
CHCHCHCCH|
CH
Sol.
313
1
3
13
2
24
3
1
3
1
CHHC
||HCHCHCCHC
|HC
o
o
ooooo
o
1° Carbon atoms = 5 , 2° Carbon atom = 1 , 3° Carbon atom = 1, 4° Carbon atom = 1
7. IUPAC SYSTEM OF NOMENCLATUREThe basic criteria for naming a structure by IUPAC system is choice of a parent name or the basic carbon
skeleton.
Nomenclature of alkanes is fundamental to naming whole class of organic compounds because it helps us
to identify the basic carbon skeleton.
General rules for IUPAC nomenclature
The IUPAC system is the most rational and widely used system of nomenclature in organic chemistry. The
most important feature of this system is that any given molecular structure has only one IUPAC name and
any given IUPAC name denotes only one molecular structure.
The IUPAC name of any organic compound consists of five parts, i.e.,
1. Word root 2. Primary suffix 3. Secondary suffix 4. Primary prefix 5. Secondary prefix
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Thus, a complete IUPAC name of an organic compound consists of the following parts :
Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
1. Word root : It is the basic unit of the name. It denotes the number of carbon atoms present in the principal
chain (the longest possible continuous chain of carbon atoms including the functional group and based upon
the common names of alkanes) of the organic molecules.
Chain length Word root Chain length Word root
C1 Meth- C7 Hept(a)-
C2 Eth- C8 Oct(a)-
C3 Prop(a)- C9 Non(a)-
C4 But(a)- C10 Dec(a)-
C5 Pent(a)- C11 Undec(a)-
C6 Hex(a)- C12 Dodec(a)-
2. Primary Suffix. A primary suffix is always added to the word root to indicate whether the carbon chain is
saturated or unsaturated. The three basic primary suffixes are given below :
Type of carbon chain Primary suffix General name
(a) Saturated – ane Alkane
(b) Unsaturated with one double bond – ene Alkene
(c) Unsaturated with one triple bond – yne Alkyne
If the parent carbon chain contains two, three or more double or triple bonds, numerical prefix such as di (for
two), tri (for three), tetra (for four) etc. are added to the primary suffix. For example,
Type of carbon chain Primary suffix General name
(a) Unsaturated with two double bonds – diene Alkadiene
(b) Unsaturated with two triple bonds – diyne Alkadiyne
3. Secondary suffix. A secondary suffix is then added to the primary suffix to indicate the nature of the
functional group present in the organic compounds. Secondary suffix of some important functional groups
are given below :
Class of organic compounds Functional group Secondary suffix
Alcohols – OH – ol
Aldehydes – CHO – al
Ketones > C = O – one
Carboxylic acids – COOH – oic acid
Acid amides – CONH2 – amide
Acid halide – COX – oyl halide
Esters – COOR alkanoate
Nitriles – CN - nitrile
Thioalcohols – SH - thiol
Amines – NH2 - amine
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The following examples illustrate the use of word root, primary suffix and secondary suffix in naming of
organic compounds.
Organic compounds Word root Primary suffix Secondary suffix IUPAC name
CH3CH2OH Eth an(e) ol Ethanol
CH3CH2CH2NH2 Prop an(e) amine Propanamine
CH3CH2CH2COOH But an(e) oic acid Butanoic acid
CH3CH2CN Prop an(e) nitrile Propanenitrile
CH2 = CHCHO Prop en(e) al Propenal
HC CCOOH Prop yn(e) oic acid Propynoic acid
4. Primary prefix. A primary prefix is used simply to distinguish cyclic from acyclic compounds.
For example, in case of carbocyclic compounds, (cyclic compounds containing only carbon atoms in the
ring), a primary prefix, cyclo is used immediately before the word root. Thus.
e.g.CH2
CH2
CH2
CH2
CH2
Cyclo
Primary prefix
+ pent
Word root
+ ane
Primary suffix
= Cyclopentane
IUPAC name
If the prefix cyclo is not used, it simply indicates that the compound is acyclic or open chain.
5. Secondary prefix. In IUPAC system of nomenclature, certain groups are not considred as functional
groups but are treated as substituents. These are called secondary prefixes and are added immediately
before the word root (or the primary prefix in case of carbocyclic compounds) in alphabetical order to denote
the side chains or substituent groups. The secondary prefixes for some groups which are always treated as
substituent groups (regardless of the fact whether the organic compound is monofunctional or polyfunctional)
are given below :
Substituent group Secondary prefix Substituent group Secondary prefix
– F Fluoro – OCH3 (– OMe) Methoxy
– Cl Chloro – OC2H5(–OEt) Ethoxy
– Br Bromo – R Alkyl
– I Iodo – CH3 (– Me) Methyl
– NO2 Nitro – C2H5 (– Et) Ethyl
– NO Nitroso – CH2CH2CH3(n-Pr) n-Propyl
Diazo – CH(CH3)2 (– iPr) Isopropyl
– OR Alkoxy – C(CH3)3 (t-But) t-Butyl
NN
Organic compoundsSecondary
prefixWord root
Primary
suffixIUPAC name
CH3CH2 – Br Bromo eth ane Bromoethane
CH3 – NO2 Nitro meth ane Nitromethane
C2H5 – OC2H5 Ethoxy eth ane Ethoxyethane
In case of carbocylic compounds, primary prefixes are also used.
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e.g. H C2
H C2
CH2
CH2
CH|Br
CH|OH
4
5
6 2
1
34-Bromo
Secondaryprefix
+ cyclo
Primaryprefix
+ hex
Wordroot
+ an (e)
Primarysuffix
+ 1-ol
Secondarysuffix
Ex.3 Write IUPAC name of following compound
SO H3
I
CH3
Sol.
SO H3
I1
CH3
2
34
5
2 – Iodo – 3 – methylcyclopentanesulfonic acid.
Here Secondary prefix = 2 – Iodo – 3 – methyl
Primary prefix = cyclo
Word root = pent
Primary suffix = ane
Secondary suffix = sulfonic acid
Ex.4 Write the IUPAC name of following compound
CH3
CH
OH
CH3 CH3
Sol. 2 – Isopropyl – 5 – methylcyclohexanol
Here Secondary prefix = 2 – Isopropyl – 5 – methyl
Primary prefix = cyclo
Word root = hex
Primary suffix = an(e)
Secondary suffix = ol
8. IUPAC Nomenclature of Branched / Complex Alkanes(1) (a) Select the longest chain of carbon atoms in the molecule.
(b) Count the number of carbon atoms in that chain and name according to the following rules.
e.g.
Longest chain has 7 carbons.
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It is a suffixprimary
ane
rootword
hept
When chains of equal lengths are competing for selection, that chain is selected which has more number ofsubstituents.
Here the chain shown is selected.
(2) Carbon atoms in the longest chain selected as above is numbered consecutively from one end to theother such that the substituents attached get the lowest number.In the above example, according to this rule, the numbering will be done as :
By this numbering, locant (substituents) get the number 2, 3 and 4 compared to 4, 5 and 6 if numbering isdone from other end.
(3) Each substituent, which, obviously, is an alkyl group is named according to number of carbon atomspresent in it and it is prefixed by the number to which it is located in the main chain.
In the above example, substituents are as follow :
–CH3(methyl) group at carbon No. 2 2–methyl
–C2H
5(ethyl) group at carbon no. 3 3–ethyl
–CH2CH
2CH
3(propyl) group at carbon no. 4 4–propyl
Hence, the above compound is named as :3–Ethyl–2–methyl–4–propylheptane
(4) If the same substituent occurs more than once in the molecule, the prefix di (for two), tri (for three), etc.are used to indicate how many times it appears.
The above example can be written with a little modification as :
e.g.
Methyl at No. 2,3, Ethyl at no. 3, Propyl at no. 4This will be named as : 3–Ethyl–2,3–dimethyl–4–propylheptane
(5) The name of the compound is composed in such a manner that each substituent with its number andname is written alphabetically just before the word root. Prefixes di, tri, tetra etc. are not considered indeciding alphabetical order.
Ethyl will be written before methyl which will be written before propyl.Note that in the above examples, this pattern has been compiled with.
Also, as per convention,(1) numbers are separated each other by commas(,).(2) numbers are separated from words by hyphens and(3) write the name of the compound as a single word (with no space between)
e.g. Write the IUPAC name of
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1. Primary suffix is ane as all are single bonds
2. Chain is numbered as shown
3. Root word is hex
4. Prefixes methyl appears twice
it is 2, 4–dimethyl and 3 – ethyl
5. While arranging in alphabetical order Replicators di, tri, tetra are not considered.
3 – Ethyl–2, 4–dimethylhexane
Ex. 5 Write the IUPAC name of following compounds
(i)
(ii)
Ans. (i) 2,2,3–Trimethylpentane (ii) 5–(1, 2 – Dimethylpropyl)nonane
Ex.6 Write IUPAC name of the following compounds:-
(a)
(b)
(c)
Ans. (a) 5-Ethyl-3-methyloctane (b) 4-Ethyl-2,2,6-trimethylheptane (c) 3-Methylhexane
9. IUPAC Nomenclature of AlkenesFunctional group : –C=C–
(1) Select the longest chain containing carbon–carbon double bond. This need not be the longest chain in the
compound as a whole. Parent name will be alkene corresponding to number of carbon atoms in the longest
chain.
e.g.
Longest chain is as shown above. It has 6 atoms hexene = parent name
(2) Carbon atoms in the longest chain is numbered so that doubly bonded carbon atom gets the lowest
number. The position of double bond is indicated by the smaller of the numbers assigned to two carbon
atoms of double bond.
The above example can be rewritten as,
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Position of double bond will be indicated as no. 1.
Hence, name will be, 3–Ethyl–3–methylhex–1–ene
e.g.
2, 2, 5, 5-Tetramethylhex-3-ene
Ex.-7 Write the IUPAC name of the following compounds:-
(a) CH3
– CH2
– – CH = CH – CH2
– CH3
(b)
CH3CH3CH – CH2 3
CH – CH3
––
––
–– –
CH – CH – CH – CH – CH – CH = C – CH3 2 3
CH2
CH2
CH3
Ans. (a) 5-Methylhept-3-ene
(b) 5-Ethyl-2,6-dimethyl-4-(3-methylbutyl)oct-2-ene
Ex.8 Draw the bond line structures of the following compounds.
(a) 2-Methylhept-3-ene
(b) 2, 6–Dimethylhepta – 1, 5–diene
Ans. (a) (b)
10. IUPAC Nomenclature of alkynes (–C C– group)Numbering of longest chain is exactly same as that for naming alkenes.
e.g.
4,4–Dimethylpent–1–yne
Ex.9 Write the IUPAC name of the following compounds:-
(a) CH3– – C C – CH
3(b)
Ans. (a) 4-Methylpent-2-yne (b) 4-Propylhept-2-yne
Ex.10 Write the IUPAC name of the following compounds:-
(a) 233 )CH(CCHC–CH
(b)
Sol. (a) 4-Methyl-pent-2yne (b) 3,4,4-Trimethylhex-1-yne
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11. IUPAC nomenclature of hydrocarbons containing both double and triple bondsoccuring only once(i) Such hydrocarbon is named as alkenyne (not alkynene).(ii) Numbering is done in a manner that double and triple bonds get the lowest possible number. If doublebond and triple bond both have same number then double bond is given preference our triple bond.
HC C – CH2
– CH = CH2
(wrong) 1 2 3 4 5 (numbering is done from alkyne)(write) 5 4 3 2 1 (numbering is done from alkene)
Ex.11 Write IUPAC name of the following compounds.
(a) (b)
Ans. (a) 3-(2-Methyl propyl)hept-1-en-4-yne (b) Oct-1-en-4-yne
Ex.12 Write IUPAC name of the following compounds.
(a) 22 CHCHCCHHC (b)
Ans. (a) Pent-1-en-4-yne (b) Hepta-3,6-dien-1-yne
12. IUPAC Nomenclature of Alicyclic Compounds
1. The names of alicyclic compounds are obtained by adding the prefix "cyclo"
e.g.
Cyclobutane Cyclopentene
2. The numbering of the carbon atoms in the ring is done in such a way that the substituent which comes firstin the alphabetical order is given the lowest possible number and it does not violate the lowest set of locantsrule.
e.g.
CH CH2 3
CH3
6
1
2
3
4
5
C H2 56
1
2
3
4
5
CH3
CH3
1 – Ethyl – 3 – methyl 2 – Ethyl – 1, 4 – dimethylcyclohexane cyclohexane
C H2 5
CH3
CH3
Br
Cl
I
1
23
4
5
6
3 – Ethyl – 1, 1 – dimethyl 2 – Bromo – 1 – chlorocyclohexane – 3 – iodocyclohexane
(3) When the ring contains more or equal number of carbon atoms than the alkyl group attached to it, then itis named as a derivative of cycloalkane and the alkyl group is treated as substituent
e.g.
CH – CH – CH2 2 3(CH ) CH2 5 3
1
2
3
4
5
6
(CH ) CH2 5 3CH (CH )3 2 5
Propylcyclopropane 1, 3, 5 Trihexylcyclohexane Cyclohexylcyclohexane
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(4) The alkane chain contains greater number of carbon atoms than present in the ring, the compound isconsidered as the derivative of alkane and the ring is designated as substituent.
e.g.
1 23
4CH – CH – CH – CH – CH – CH3 2 2 2 3
2 – Cyclopropylbutane 3 – Cyclopentylhexane
(5) If ring has unsaturation and side chain is saturated then ring is selected as parent chain.If side chain has unsaturation and ring is saturated then side chain is selected as parent chain.If both have unsaturation the chain with maximum unsaturation has selected as parent chain.If equal unsaturation then longest chain is selected as parent chain.If unsaturation and number of carbon atoms both are equal then ring is selected as parent chain.
e.g.
1
2
3
4
5
6
1
234
5
6
1 – Ethylcyclohex – 1 – ene 6 – Ethyl – 3, 3 – dimethylcyclohex – 1 – ene
1
23
CH2
123CH – CH = CH2 2
Methylenecyclohexane 3 – Cyclopropylprop – 1 – ene
1–(Hex–3–enyl)cyclohex–1–ene
(6) If more than one alicyclic ring is attached to a single chain, the compound is named as a derivative ofalkane and the ring is treated as a substituent group.
e.g. CH2
Dicyclopropylmethane 1 – Cyclohexyl – 4 – cyclopropylbutane
(7) If a multiple bond and some other substitutents are present in the ring, the numbering is done in such away that the multiple bond gets the lowest number
e.g.2
3
4
5
6
1
NO2
3 – Nitrocyclohex – 1 – ene
(8) If a compound contains an alicyclic ring directly linked to the benzene ring. It is named as a derivative ofbenzene.
e.g.
CH3
NO21
2 3
4
56
1
23
4
5 6
Cyclohexylbenzene 1 – (2 – Methylcyclohexyl) – 4 – nitrobenzene
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CHEMISTRY
(9) If functional group is present in cyclic compounds than the main chain is taken in which principal
functional lie's, if the principal functional group is present in ring also then main chain will be taken for the
maximum no. of carbon atoms.
e.g.
OH
2
1
6
5
4
3 OH
231
2 – Propylcyclohexan – 1 – ol 1 – Cyclohexylpropan – 2 – ol
(10) When chain terminating functional group is directly attached with ring then ring is taken as parent chain
& special suffix used for functional group.
Functional Group Suffix
CHO Carbaldehyde
COOH Carboxylic Acid
COX Carbonyl halide
COOR Alkyl Carboxylate
CONH2 Carboxamide
CN Carbonitrile
e.g.
CN CHO
Cyclohexanecarbonitrile Cyclohexanecarbaldehyde
O
COOC H2 5
1
2
Ethyl (2–oxo)cyclohexane – 1 – carboxylate
Ex.13 Write the IUPAC name of following compound
(A)
CH = CH – COCH3
(B) CH3
CH3
Sol. (A)
CH = CH – C – CH3
O4
3 2 1
(B)
4 – Cyclohexylbut – 3 – en– 2-one 1,5–Dimethylcyclopent–1–ene
Ex.14 The correct IUPAC name of following compound is
CH – CH – CH – CH2 2 3
OHOH
Sol. 2 – (2 – hydroxybutyl) cyclohexan – 1 – ol
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13. IUPAC Nomenclature of Compounds Containing Functional Groups
Functional groups
Non chainterminating
Chainterminating
(i)
– C –
– OH
– NH2
– SH
– C – OH
– C – O – C –
– C – OR
– C – X
– C – NH2
– CN
– CHO
||O
|| ||
||
||
||
O O
O
O
O
||O
– SO H3
(ii)
(iii)
(iv)
(v)
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
13.1 Rules for non chain terminating functional groups(1) Parent chian : Select the longest possible chain with maximum functional group and maximum unsaturation
without caring whether it also denotes the longest possible chain or not.
e.g. CH – CH – CH – CH – CH3 2 2 3
|CH OH2
4 3
1
2
(Parent chain contains four rather than five carbon atoms)
(2) Lowest number for the functional group
Numbering is done from that side of the chain which gives lowest locant to the principle functional group
followed by double and triple bonds.
e.g. CH – CH – CH – C – CH – CH3 2 2 3
|CH3
6 5 34 ||O
2 1
( ) correctI
CH – CH – CH – C – CH – CH3 2 2 3
|CH3
1 2 43 ||O
5 6
( ) wrongII
( C = O group gets lowest number 3) ( C = O group gets number 4 which is not lowest)
13.2 Rules for chain terminating functional groups(1) When a chain terminating functional group such as – CHO, – COOH, –COOR, –CONH
2, –COCl, –CN
etc. is present, it is always given number 1 (one.)
e.g. CH – CH – CH – CH3 2 3
|COOH
4 3 2
1
2-Methylbutan-1-oic acid
CH – C C – CH – C – H3 2||O
5 4 3 2 1
Pent-3-yn-1-al
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(2) If a compound contains two or more like groups, the numerical prefixes di, tri, tetra etc. are used
e.g.
triol3,2,1opanePr
OHOHOH|||CHCHCH 22
CH – C – CH – C – CH3 2 3
|| ||O O
Pentane-2, 4-dione
(3) The name for benzene as substituent is phenyl. In case the phenyl ring is further substitued, the carbon
atoms of the ring directly attached to the parent chain in such a ways that the substituent on the ring gets the
least possible number. For example
(4) If the organic molecule contain more than one similar complexes substitutents, then the numerical
prefixes such as di, tri, tetra etc. are replaced by bis, tris, tetrakis etc. respectively.
e.g.HO – CH – CH – O2 2
HO – CH – CH – O2 2
CH – COOH
2, 2-Bis (2-hydroxyethoxy) ethanoic acid
Ex.14 Write IUPAC Name
(a)
––
CH3
CH2
CH CH – CH – C – OCH3 2 3
O
(b) (CH3)3COH (c)
Sol. (a) Methyl-2-ethylbutanoate (b) 2-Methylpropan-2-ol (c) 3-Methylbutanoic acid
13.3 Rules for IUPAC nomenclature of polyfunctional compounds :(1) When an organic compound contains two or more different functional groups than senior functional group
is selected as the principal functional group while other groups are treated as substituents.
(2) Some functional group such as all halo groups (fluoro, bromo, chloro, iodo), nitroso (NO) nitro (– NO2) and
alkoxy (–OR) are always treated as substituent groups.
e.g. CH – CH – CH – CH – CH3 3
| |Cl OH
|NH2
3 2 1
4-Amino-3-chloropentan-2-ol(– NH & – Cl group treat as substituent)2
45
Numbering the principal chain order is
[Principal functional group > double bond > triple bond > substituents]
e.g. CH – C – CH – COOH3 2
||O
3-Oxobutan-1-oic acid
CH – C – CH – CH – CHO3 2 2
||O
4-Oxopentan-1-al
5 4 2 13
[– CHO > C=O]
[– COOH > – CO]
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O = CH – CH – CH – C – CH – COOH2 2 2
6 5 4 ||O
3 2 1
3, 6-Dioxohexanoic acid or 5-Formyl-3-oxopentanoic acid
[COOH > C & CHO]||O
(3) The longest possible chain of carbon atoms containing the functional group and the maximum number of
multiple bonds is selected as parent chain.
e.g. (a)
OHHC|
CH–CH–HC–HC–HC
21
32
2
2
3
3
4
parent chain contains four rather than five carbon atoms .
(b)
parent chain contains five rather than six carbon atoms
(c)CH – CH – CH – C – O – C H3 2 2 5
||O
|CH = CH2
123
4 5
(d)CH – C CH – CH – CH3 3
|CH3
1
234
|CH3
|C – NH2
||O
|CH3
Ethyl 3-methylpent-4-en-1-oate 3,3 Dimethyl –2 – (1–methylethyl)butanamide
(4) If more than one same chain terminating group are present then the principal chain is selected including
the functional groups and numbring is done from that side which gives lowest locant to unsaturation
substituent.
e.g (a) HOOC – CH – CH – COOH2 2
1 2 3 4
Butane-1, 4-dioic acid
(b)
edinitriletanMethylpen2CH|
NCHCHCHCCN
3
5
2
4
2
321
e.g. Write the IUPAC name of
3221
3223
CHCHCHCN
||CHCHCHCHCHCH
762
543
1. The longest chain containing functional group is of 7 carbon atoms. Therefore, the word root is hept
& the chain is numbered as shown.
2. There is no multiple bond in it. Hence, the primary suffix is ane.
3. The functional groups is – CN. Hence, secondary suffix is nitrile
4. Moreover, there is a methyl group on carbon 5 and ethyl group on carbon 3.
5. The IUPAC name is, therefore, 3-Ethyl-5-methylheptanenitrile
e.g. Write the IUPAC name of
3
223
CH|
CHOCHCHCHCHCHCH1234567
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1. The longest chain containing functional group is of seven carbon atoms. Therefore, word root is hept.2. As C = C double bond is present in the molecule. Thus, the primary suffix is ene.3. The secondary suffix is al because the presence of – CHO group.4. The chain is numbered as shown so that carbon atom of – CHO groups gets number 1.
The methyl group is present on carbon 5 while position of double bond is 3. Thus, IUPAC name is5-Methylhept-3-en-1-al.
e.g. Write the IUPAC name of
CH – CH – CH CH CH – CH – CH3 2 3 2 3
11
NH2
10
9 3 2 1
HC
Br
CH CH OH
HC CH
NO2
8 7 4
56
1. Primary suffix is ene, due to presence of double bond between C4& C
5
2. Senior functional group is alcohol hence secondary suffix is -ol3. Root word is undec4. Chain is numbered as shown5. 6–nitro–7–methyl–8–Bromo–10–aminoare prefixes.Arrangethem in alphabaeticalorder&give thename.
10 – Amino– 8 – bromo – 7 – methyl – 6 – nitroundec – 4 – en – 2 – ol
Ex.15 Write IUPAC Name
(a) Ans. 4-Oxopentanoic acid
(b) Ans. Propane-1,2,3-tricarboxylic acid
(c) Ans. 4-(But-2-enyl)non-6-enoic acid
Ex.16 Write IUPAC Name
(a) Ans. 1-Chloro-5-methylhexan-2-one
(b) Ans. 3-Chlorobutan-1-ol
(c) Ans. N-methyl-2-Methoxyethan-1-amine
(d) Ans. 3-Bromo-2-chloro-5-methyloctane
14. Nomenclature of Aromatic CompoundsThe aromatic compounds are cyclic compounds which contain one or more benzene type rings. Benzene is
the simplest hydrocarbon of aromatic series which has planar cyclic ring of six carbon atoms having threedouble bonds in alternate positions as shown below.
CH
CH
CH
CH
HC
HC
or
1
2
3
4
5
6
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(i) Nuclear Substituted –The functional group is directly attached to the benzene ring in the IUPAC system, they are named as
derivatives of benzene. The position of the substituents in disubstituted benzenes are indicated either by
prefixes such as O–(ortho) for 1, 2, m –(meta) for 1, 3 and p (para) for 1, 4 position. However, many of their
common names have also been adopted by the IUPAC system
(ii) Side – Chain Substituted –
The functional group is present in the side chain of the benzene ring in the IUPAC systems, these are usually
named as phenyl derivatives of the corresponding aliphatic compounds.
The IUPAC and common names of a few important members of each family are given below.
(1) Aromatic hydrocarbons (arenes) hydrocarbons which contain both aliphatic and aromatic units are called
arenes . These are of two types
(a) Hydrocarbon containing one ring only.
e.g.
1
2
CH3
CH3
1
2
3
CH3
CH3
1, 2 – Dimethyl benzene 1, 3 – Dimethyl benzene
(o – Xylene) (m – Xylene)
1
2
3
4
CH3
CH3
1, 4 – Dimethyl benzene Isopropylbenzene
(p – Xylene) (Cumene)
CH = CH2
123CH CH = CH2 2
Phenylethene 3 – Phenylprop-1-ene
(Styrene) (Allyl benzene)
C CH123
C CHCH2
Phenylethyne 3-Phenylprop-1-yne
(Propargyl benzene)
(b) Hydrocarbon containing condensed or fused ring
1
2
3
45
6
7
8 1
2
3
410
6
7
8 9
5
1
2
3
4
10
6
7
8 9
5
Naphthalene Anthracene (Phenanthrene)
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2. Aryl Groups :
CH2 CH C
Phenyl (Benzyl) (Benzal) (Benzo)
1CH3
2
1CH3
2
3
1CH3
2
3
4
2 – Tolyl 3 – Tolyl 4 – Tolyl
or (o – Tolyl) or (m – Tolyl) or (p – Tolyl)
3. Halogen derivatives
1Cl
1Cl
2Cl
1Cl
Cl
Chlorobenzene 1,2 – Dichlorobenzene 1, 4 – Dichlorobenzene
or o – Dichlorobezene or p – Dichlorobenzene
CHCl21
CH – CH – CH – Br2 2 2
23CH = CH – Cl
Phenyldicholoromethane 1 – Bromo – 3 – phenylpropane 1 – chloro – 2 – phenylethene
(Benzyl chloride)
4. Hydroxy derivaties
The nuclear hydroxy derivatives are called phenols while the side chain substituted hydroxy derivatives are
called aromatic alcohols.
(i) Phenols – monohydric
OH
3
OH
2
1
CH3
4
Hydroxybenzene 4 – methyl phenol
(phenol) (p –cresol)
(ii) Dihydric and polyhydric phenols
OH
1OH
2
OH
3 OH
1
2
OH
OH
Benzene – 1, 2 – diol Benzene – 1, 3 – diol Benzene – 1, 4 – diol
(Catechol) (Resorcinol) (Quinol)
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(iii) Aromatic Alcohols
CH OH2 CH – CH – OH3
2 1
Phenyl methanol 1 – phenylethanol
(Benzyl alcohol) (–phenyl ethyl alcohol)
(iv) Aromatic ethers
OCH3 O – C H6 5
Methoxy benzene Phenoxybenzene
(Anisol) (Diphenyl ether)
(v) Aldehydes
CHO CHO
OH
2
1
Benzenecarbaldehyde 2 – Hydroxybenzenecarbaldehyde
(Benzaldehyde) (Salicylaldehyde)
CH – CH – CHO2 2
3 – Phenylpropanal
( – phenylpropionaldehyde)
(vi) Ketones
2CO CH3
1COC H6 5
Methyl phenyl ketone Diphenyl ketone
(Acetophenone) (Benzophenone)
(vii) Nitro Compounds
NO2
2
1
3
NO2
NO2
Nitrobenzene 1, 3 – Dinitrobenzene
(m – Dinitrobenzene)
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1CH3
NO2
2
1
3
OH
NO2NO2
NO2
4
5
6
2 – Nitro toluene 2, 4, 6 – Trinitrophenol
(O – Nitro toluene) (T.N.P) (Picric acid)
(viii) Amines
(a) Aryl amines
NH2 NH2
NH2
Benzenamine Benzene –1, 2 – diamine
(Aniline) (O – phenylenediamine)
N(CH )3 2 NHC H6 5
N, N – Dimethylbenzenamine N – Phenyl benzenamine
(N, N – Dimethylaniline) (Diphenyl amine)
(b) Aryl alkyl amine
CH NH2 2 CH – CH – NH2 2 2
Phenylmethamine 2 – Phenyl ethanamine
(Benzylamine) ( – phenyl ethyl amine)
(ix) Arenediazonium Salts
N N.Cl N NHSO4
Benzene diazonium chloride Benzene diazonium hydrogen sulphate
(x) Cyanides and Isocyanides
C N N C
Benzenecarbonitrile Benzeneisonitrile
(Benzonitrile) (Phenyl isocyanide)
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(xi) Carboxylic Acids
COOH COOH
CH3
1
2
COOH
COOH1
2
Benzenecarboxylic acid 2 – Methylbenzene Benzene – 1, 2 -dicarboxylic acid
carboxylic acid (Phthalic acid)
(O – toluic acid)
(xii) Anhydrides :5656 HCCOCHC
||||OO
Benzenecarboxylic anhydride
(xiii) Esters :O
CH3
1
2
34
O – C – CH3
4–Methylphenyl ethanoate
(xiv) Amides :
O
C – NH2
Benzenecarboxamide
Ex.17 Write IUPAC name of following aromatic compounds
(a)
CH – CH – CHO2
|CH3
3 2 1
2-Methyl-3-phenylpropanal
(b)
15. Some Important 1993 Recommendations of IUPAC Nomenclature of Organic Compounds
1. Locants (numerals and / or letters) are placed immediately before the part of the name to which they
relate. For example.
CH3CH
2CH = CH
2should be named as but-1-ene
CH3CH
2OH should be named as ethan-1-ol
similarly, a few more examples are given as following :
2. The locant 1 is often omitted when there is no ambiguity. For example.
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In all the above examples locant 1 for the functional group is omitted because the position of the functional
group is unambiguous. However, in the following cases the position of the functional group must be mentioned.
Here, we cannot write simply propanol (or propanamine) because there are two propanols ; propan-1-ol and
propan-2-ol.
3.Arrangement of Prefixes
(i) Simple prefixes such as methyl, ethyl, chloro, nitro, hydroxy, etc. are arranged alphabetically.
The prefixes di, tri, etc. are however not considered for comparison.
e.g.CH CH2 2
12
| |BrCl
1-Bromo-2-chloroethane
(ii) The name of a prefix for a substituted substituent is considered to begin with the first letter of its complete
name.
e.g.
for the substituted 1-chloropropyl, 'C' is taken as the first letter.
(iii) When two or more prefixes consist of identical roman letters priority for citation is given to the group
which contains the lowest locant at the first point of difference.
For example,
Here, 1-chloroethyl gets priority over 2-chloroethyl.
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Compound Prefix Wordroot Pri. Suffix Sec. Suffix Name
CH3– CH
2– CH
2–CH
3– But ane – Butane
2–Methyl But ane – 2–Methylbutane
CH3–CH=CH–CH
2–CH
3– Pent 2-ene – Pent–2-ene
2–Methyl Pent 2-ene – 2-Methylpent-2-ene
CH C – CH3
– Prop yne – Propyne
– But ane 1–ol Butan–1–ol
3–Methyl But ane 1–ol 3–Methylbutan–1–ol
4 – Methyl Pent 3–ene 1–ol 4-Methylpent-3-en-1-ol
– Prop 2–yne 1–ol Prop–2–yn–1–ol
CH C – C CH – Buta diyne – Butadiyne
CH3–CH=C=CH–CH
3– Penta 2, 3 diene – Penta-2, 3–diene
Cyclo But 2-ene 1-ol Cyclobut-2-en-1-ol
4-Nitro Benz – oyl Bromide
16. Additional Illustrations and Solved Examples
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17. Word roots
Number of
Carbon AtomsWord Root Name
1 Meth Methane
2 Eth Ethane
3 Prop Propane
4 But Butane
5 Pent Pentane
6 Hex Hexane
7 Hept Heptane
8 Oct Octane
9 Non Nonane
10 Dec Decane
11 Undec Undecane
12 Dodec Dodecane
13 Tridec Tridecane
14 Tetradec Tetradecane
15 Pentadec Pentadecane
16 Hexadeca Hexadecane
17 Heptadec Heptadecane
18 Octadec Octadecane
19 Nonadec Nonadecane
20 Icos Icosane
21 Heneicos Heneicosane
22 Docos Docosane
23 Tricos Tricosane
30 Triacont Triacontane
31 Hentriacont Hentriacontane
40 Tetracont Tetracontane
50 Pentacont Pentacontane
60 Hexacont Hexacontane
70 Heptacont Heptacontane
80 Octacont Octacontane
90 Nonacont Nonacontane
100 Cent & Hect Centane or Hectane
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Class
R – COOH
R – SO3H
R – C – O – C – R|| ||O O
R – COOR
R – C – X||O
R – C – NH2
||O
R – C N
R – C – H||O
R – C – R||O
R – OH
R – SH
R – NH2
Name
Alkanoic Acid
Alkane sulphonic Acid
Alkanoic Anhydride
Alkyl alkanoate
Alkanoyl halide
Alkanamide
Alkanenitrile
Alkanal
Alkanone
Alkanol
Alkanethiol
Alkanamine
Suffix
– oic Acid (carboxylic acid)
– sulphonic Acid
– anhydride
– alkanoate (carboxylate)
– oyl halide (carbonyl halide)
– amide (carboxamide)
– nitrile (carbonitrile)
– al (carbaldehyde)
– one
– ol
– thiol
–amine
Prefix
Carboxy
sulpho
------------
Alkoxy carbonyl
halo carbonyl
Carbamoyl
cyano
formyl / oxo
Oxo / Keto
hydroxy
mercapto
amino
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
18. Functional Groups in decreasing order of seniority
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19. Common and IUPAC Names of Some Organic Compounds
S.No. Compound Common names IUPAC name
1. CH4
Methane or Marsh gas Methaneor Fire damp orCarbane
2.
3
33
CH|
CHCHCH Isobutane 2-Methylpropane
3.
3
323
CH|
CHCHCHCH Isopentane 2-Methylbutane
4.
3
33
3
CH|
CHCCH|CH
Neopentane 2, 2-Dimethylpropane
5. 3223
3
CHCHCHCHCH|CH
Isohexane 2-Methylpentane
6.
3
323
3
CH|
CHCHCCH|
CH
Neohexane 2-2-Dimethylbutane
7. CH2= CH
2Ethylene Ethene
8. CH3CH = CH
2Propylene Propene
9. CH3CH
2– CH = CH
2-Butylene But-1-ene
10. CH3CH = CHCH
3-Butylene But-2-ene
11. CH3(CH
2)2CH = CHCH
3-Hexylene Hex-2-ene
12. Isobutylene 2-Methylpropene
13. CH2
= C = CH2
Allene Propadiene
14. HC CH Acetylene Ethyne
15. H3C – C CH Methyl acetylene Propyne
16. CH3– Cl Methyl chloride Chloromethane
17.
I|
CHCHCH 33 Isopropyl iodide 2-Iodopropane
18.
Cl|
CHCHCHCH 323 sec-Butyl chloride 2-Chlorobutane
19. ClCHCHCH|CH
23
3
Isobutyl chloride 1-Chloro-2-methylpropane
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S.No. Compound Common name IUPAC name
20.
3
3
3
CH|
ClCCH|
CH
tert-Butyl chloride 2-Chloro-2-methylpropane
21.
BrCH|
BrCH
2
2
Ethylene dibromide 1, 2-Dibromoethane
22.
Br|
BrCHCH3 Ethylidene bromide 1, 1-Dibromoethane
23. CH2
= CH – Cl Vinyl chloride Chloroethene
24. CH2
= CH – CH2– Br Allyl bromide 3-Bromo-1-propene
25. CH3– CH
2– OH )inolMethylcarb(
oralcoholEthylEthanol
26. CH3CH
2CH
2– OH )carbinolEthyl(
oralcoholopylPrn Propan-1-ol
27.
OH|
CHCHCH 33 Isopropyl alcohol Propan-2-ol
28. CH2= CH – CH
2– OH Allyl alcohol Prop-2-en-1-ol
29. CH C – CH2– OH Propargyl alcohol Prop-2-yn-1-ol
30.OHOH||CHCHCH 23
Propylene glycol Propane-1, 2-diol
31. HO – CH2– CH
2– CH
2– OH Trimethylene glycol Propane-1, 3-diol
32.
OHOHOH|||
CHCHCH 22 Glycerol or Glycerine Propane-1, 2, 3-triol
33. H – CHO Formaldehyde Methanal
34. CH3– CHO Acetaldehyde Ethanal
35. CH3CH
2CH
2– CHO n-Butyraldehyde Butanal
36. Isobutyraldehyde 2-Methylpropanal
37. CH2= CH – CHO Acrolein Propenal
38. CH3CH = CH – CHO Crotonaldehyde But-2-enal
39. CH3– CO – CH
3Dimethyl ketone or Acetone Propanone
40. CH3– CO – CH
2CH
3Ethyl methyl ketone Butanone
41. CH3– CO – CH
2CH
2CH
3Methyl n-propyl ketone Pentan-2-one
42. CH3CH
2– CO – CH
2CH
3Diethyl ketone Pentan-3-one
43. CH3CO – CH = CH
2Methylvinyl ketone But-3-en-2-one
44. H – COOH Formic acid Methanoic acid
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S.No. Compound Common name IUPAC name
45. CH3– COOH Acetic acid Ethanoic acid
46. CH3CH
2CH
2– COOH n-Butyric acid Butanoic acid
47. CH3CH
2CH
2CH
2COOH n-Valeric acid Pentanoic acid
48. Iso-butyric acid 2-Methylpropanoic acid
49. CH2= CH – COOH Acrylic acid Propenoic acid
50.
COOH|COOH Oxalic acid Ethanedioic acid
51. Malonic acid Propanedioic acid
52.
COOHCH|
COOHCH
2
2
succinic acid Butanedioic acid
53. Glutaric acid Pentanedioic acid
54.
OH|
COOHCCH|H
3 Lactic acid 2-Hydroxypropanoic acid
55.COOHCCH
||O
3 Pyruvic acid 2-Oxopropanoic acid
56.
HOCHCOOH|
HOCHCOOH Tartaric acidaciddioic
etanuDihydroxyb3,2
57. Citric acid acidlictricarboxy3,2,1paneHydroxypro2
58.
COOHCH|
COOHCHHO
2
Malic acidaciddioic
etanbuHydroxy2
59. Maleic acid cis-But-2-enedioic acid
60. Fumaric acid trans-But-2-enedioic acid
61. CH2= CH – COOH Acrylic acid Propenoic acid
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62. H3C – CH = CH – COOH Crotonic acid But-2-enoic acid
63. H – COOCH3
Methyl formate Methyl methanoate
64. H – COOC2H
5Ethyl formate Ethyl methanoate
65. CH3– COOC
2H
5Ethyl acetate Ethyl ethanoate
66. H – COCl (unstable) Formyl chloride Methanoyl chloride
67. CH3– COCl Acetyl chloride Ethanoyl chloride
68. (CH3CO)
2O Acetic anhydride Ethanoic anhydride
69. (CH3CH
2CO)
2O Propionic anhydride Propanoic anhydride
70. H – CONH2
Formamide Methanamide
71. CH3– CONH
2Acetamide Ethanamide
72. CH3– CH
2– CONH
2Propionamide Propanamide
73. CH3
– O – N = O Methyl nitrite Methyl nitrite
74. CH3CH
2– O – N = O Ethyl nitrite Ethyl nitrite
75. CH3– NH
2Methylamine Methanamine
76. (CH3CH
2)2NH Diethylamine N-Ethylethanamine
77. (CH3)3N Trimethylamine or N,N-Dimethyl methan-
N,N-Dimethylamino- aminemethane
78. H2N – SO
3H Sulphamic acid Aminosulphonic acid
79. CH3– CN Methyl cyanide or Ethanenitrile
Acetonitrile
80. CH3
– N+ C¯ Methyl isocyanide or Methane isocyanideMethyl carbylamine
81. CH3CH
2– N+ C¯ Ethyl isocyanide or Ethaneisocyanide
Ethyl carbylamine
82. Dioxane 1, 4-Dioxacyclohexane
83. Trioxane 1,3,5-Trioxacyclohexane
84. Toluene Methylbenzene or Toluene
85. Xylene (o,m,p) (o,m,p) Dimethylbenzene
86. Mesitylenebenzene
Trimethyl5,3,1
S.No. Compound Common name IUPAC name
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87. Cumene Isopropylbenzene
88. Benzyl chloride Chlorophenylmethane
89. Benzal chloride Dichlorophenylmethane
90. Benzochloride Trichlorophenylmethane
91. Benzonitrile Benzenecarbonitrile
92.
BHC
)666orLindaneorGammexane( ecyclohexanHexachloro
[Benzene hexachloride]]
93. Carbolic acid Phenol
94. Catechol Benzene-1,2-diol
95. Resorcinol Benzene-1,3-diol
96. Hydroquinone Benzene-1,4-diol
S.No. Compound Common name IUPAC name
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97. Oil of bitter almonds Benzenecarbaldehyde
98. Salicylaldehyde
)decarbaldehyneHydrobenze2(
debenzaldehyHydroxy2
99. Phthalaldeyde
100. Acetophenone ketoneylMethylphenorneAcetopheno
101. BennzophenoneBenzophenone(Diphenylketone)
102. Phenacyl chloride Chloroacetophenone
103. Cinnamic acid acidoicen2Phenylprop3
104. )acidsalicylicAcetyl(
Aspirin
105. )salicylateMethyl(greenerintwofOil
boxylatebenzenecarhydroxy2Methyl
106. Oil of mirbane Nitrobenzene
107. Picric acid 2, 4, 6-Trinitrophenol
S.No. Compound Common name IUPAC name
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108. Acetanilide N-Phenylethanamide
109. StyrenezeneEthenylben
orethenePhenyl
110. Anisole Methoxybenzene
111. Phenetol Ethoxybenzene
112. Aniline Aniline (Benzenamine)
113. (o, m, p) Toluidine Methylaniline
114. (o, m, p) cresol Methylphenol
115. Napthalene Naphthalene
116. Anthracene Anthracene
117. Phenanthrene Phenanthrene
118. Pyrene Pyrene
S.No. Compound Common name IUPAC name
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Isomerism
1. DEFINATION OF ISOMERISM :When two or more compounds have same molecular formula but they differ in their physical properties orchemical properties or both then they are called isomers & the phenomena is called isomerism. Isomers arealso sometimes called as isomerides.
Classification of Isomerism
2. STRUCTURAL ISOMERISM :When two or more organic compounds have same molecular formula but different structural formula,(i.e., they differ in connectivity of atoms) then they are called structural isomers and the phenomena iscalled structural isomerismStructural isomers have always different IUPAC name
Ex.1ene2But
CHCHCHCH 33
, Isomersstructural
ene1ButCHCHCHCH 223
Ex.2
etanbuMethyl2
CH
|
CHCHCH
|
CH
3
32
3
,
)isomersnot(
compoundsIdentical
etanMethylbu23
323
CH|
CHCHCHCH
Isomers Characteristics Conditions
(1) Chain IsomersThey have different size ofmain chain or side chain
They have same nature oflocants
(2) Positional IsomersThey have different position of
locants
They should have same size ofmain chain and side chain and
same nature of locant
(3) Functional Isomers Different nature of locantChain and positional isomerism is
not considered
(4) MetamerismDifferent nature of alkyl groupalong a polyvalent functional
group
They should have same nature offunctional groups chain &
positional isomer is ignored
(5) TautomerismDifferent position of hydrogenatoms
The two functional isomersremains in dynamic equilibrium
to each other
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2.1 Chain Isomerism or nuclear isomerism or skeletal isomerismChain isomers have the same molecular formula but differ in the order in which carbon atoms are bonded to
each other.This type of isomerism is due to difference in the arrangement of carbon atoms i.e. straight or
branched chain of carbon atoms. It is also known as nuclear or skeletal isomerism
Condition : – 1. Size of main chain or side chain or both should be different.
2. Nature of functional group, multiple bond or substituent should not change.
Ex.3 (A) CH3– CH
2CH
2– CH
3
Butane 2–Methylpropane
Size of main chain = 4 Size of main chain = 3
Size of side chain = 0 Size of side chain = 1
Isomeric pair :
When more than one structure can be written with a given molecular formula, each of the structure with
respect to any other structure is called an isomeric pair.
Isomerism between two structures 'a1' & 'a
2' can also be seen as a relation a
1Ra
2between them.
Ex.4
)a(
CHCHCHCHCH
1
32223 ,
)a(
CH
|
CHCHCHCH
2
3
323 ,
)a(
CH
|
CHCCH
|
CH
3
3
33
3
a1R a
2 CI (chain isomers)
a1R a
3 CI
a2R a
3 CI
Ex.5 C6H
14= 5 isomers
)a(
CHCHCHCHCHCH
1
322223
,
)a(
HC
|
CHCHCHCHCH
2
3
3223
,
)a(
CH|
CHCHCHCHCH
3
3
3223
)a(
CH
|
CHCHCHCH
|
CH
4
3
323
3
)a(
CHCH||
CHCHCHCH
5
33
33
a1R a
2 CI
a1R a
3 CI
a1R a
4 CI
a1R a
5 CI
a2R a
3 not CI
a2R a
4 CI
a2R a
5 CI
a3R a
4 CI
a3R a
5 CI
a4R a
5 not CI
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No. of alkane isomers as a function of no. of carbon atoms :
no. of
carbon atomsmolecular formula no. of isomers
1 CH4 1
2 C2H6 1
3 C3H8 1
4 C4H10 2
5 C5H12 3
6 C6H14 5
7 C7H16 9
8 C8H18 15
9 C9H20 35
10 C10H22 75
15 C15H32 4347
20 C20H42 36,719
Chain isomerism in alicyclic compounds :Ex.6 C
8H
16
Size of main chain = 6 Size of main chain = 6Size of side chain = 2 Size of side chain 1 = 1
Size of side chain 2 = 1
Ex.7
Size of main chain = 6 Size of main chain = 3Size of side chain = 0 Size of side chain 1 = 1
Size of side chain 2 = 1Size of side chain 3 = 1
Ex.8
a1R (a
2– a
5) C.I.
a2R (a
3– a
5) not C.I.
a3R (a
2, a
4, a
5) not C.I.
a4
R (a2
, a3, a
5) not C.I.
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Chain isomerism in aromatic compounds
(1) (2) (3) (4)
(1) R (2) C.I.(1) R (3) C.I.(1) R (4) C.I.(2) R (3) C.I.(2) R (4) C.I.(3) R (4) not C.I.
Ex.9 Molecular formula is C5H
12
(i) CH3– CH
2– CH
2– CH
2– CH
3(ii)
3
323
CH
|
CHCHCHCH
Pentane 2 – Methyl butaneSize of main chain = 5 Size of main chain = 4Size of side chain = 0 Size of side chain = 1
(iii)
3
33
3
CH|
CHCCH|
CH
2, 2 – Dimethyl propaneSize of main chain = 3Size of side chain = 2
Compound (i) and (ii), (i) and (iii), (ii) and (iii) are chain isomers
2.2 Position isomerism :When two or more organic compounds have same molecular formula but they differ in the position of thefunctional group or the substituent, then they are called position isomers & the phenomena is called positionisomerism.Condition : Same size of main chain or side chain but different position of group or substituents or functionalgroup.
Ex.10 (a) isomerspositionCHCHCHCHCHCHCHCH
)ene2but(
)ene1but(
23
223
(b) isomerspositionCHCHCCCHCHCHCHCHC
)yne2pent(
)yne1pent(
323
322
(c)
Cl|
isomerspositionCHCHCH
ClCHCHCH
chlorideIsopropyl)anechloroprop2(
)anechloroprop1(
33
223
(d) Cl – CH = CH – Cl ] position isomers
(dichloro ethene) (dichloroethene)or (ethylidene chloride) or (ethylene chloride)
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(e) C8H
10(3-aromatic position isomers)
(I)
o-xylene
(II)
m-xylene
(III)
p-xylene
position isomers
(f) C7H
7NO
2(3-aromatic position isomers)
(I) (II) (III)
(I) R (II) P. I.(I) R (III) P. I.(II) R (III) P. I.
(g) C6H
3(OH)
3 (3 aromatic position isomers)
(I) (II) (III)
2.3 Functional IsomerismCompound having same molecular formula but different functional groups in their molecules show functionalisomerism and are called functional isomers.Condition : Functional group should be different, chain and position isomerism are not considered.
Formula Functional Isomers
(i) CnH2n Alkene, Cycloalkane
(ii) CnH2n-2 Alkadiene, Alkyne, Cycloalkene
(iii) CnH2n+2O Alcohol, Ether
(iv) CnH2nO Aldehyde, Ketone, Oxirane (epoxy)
(v) CnH2nO2
Acid, ester, hydroxy, carbonyl
(aldehyde, ketone compounds)
Ex.11 CH3– CH
2OH CH
3– O – CH
3
Ethanol MethoxymethaneFunctional groups – OH Function groups – O –
Functional groups – COOH Functional groups
Ex.12 What type of isomerism exist between four structural isomers of C4H
9NH
2
)(
NH|
CHCHCHCH
2
323
I
)(
CH
|
NHCHCHCH
3
223
II
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CHEMISTRY
)(
CH
|
NHCCH
|
CH
3
23
3
III
)V(
NHCHCHCHCH 22223
I
Ans. (I) and (II) are chain isomers (I) and (III) are chain isomers(I) and (IV) are position isomers (II) and (III) are position isomers(II) and (IV) are chain isomers (III) and (IV) are chain isomers
Ex.13 Functional isomers of CH3
– C CH is
(A) CH2
= C = CH2
(B) (C) (D) Both A and B
Ans. D (Hint : sp carbon not possible in C3–C
7rings.)
2.4 Ring chain isomers :When two or more organic compounds have same molecular formula but they differ in their structures due toone being a cyclic compound & the other one an open chain compound, then they are called ring chainisomers & the phenomena is called ring-chain isomerism or functional isomerisum.
Ex.14 Write all isomers of C4H
8and indicate relationship between them :
C4H
8
(1) R(2) R.C.I. (2) R(3) R.C.I (3) R(5) CI(1) R(3) P.I. (2) R(4) C.I. (4) R(5) RCI(1) R(4) R.C.I. (2) R(5) RCI(1) R(5) C.I. (3) R(4) R.C.I
Ex.15 Compound CH3– CH
2– CH
2– C CH and show..
(A) Position isomers (B) Chain isomers(C) Ring chain isomers (D) All of these
Ans. C
Ex.16 1, 2–Epoxy propane and prop–2–en–1–ol (Allyl alcohol) are(A) Position isomers (B) Functional isomers(C) Ring chain isomers (D) Both B and C
Ans. (D)
2.5 MetamerismThis type of isomerism is due to the difference in the nature of alkyl groups attached to the polyvalent atom
or functional group. Metamers belong to the same homologous series. Compounds like ethers, thioethers,secondary amines etc. show metamerism.
Condition : 1. Functional group should be polyvalent, hetroatomic.2. Nature of functional group should not change3. Chain and position isomerism is not considered.
Ex. 17 C2H
5– O – C
2H
5C
3H
7– O – CH
3
Diethyl ether Methyl propyl ether
Hydrocarbon groups – C2H
5, – C
2H
5Hydrocarbon groups – C
3H
7, – CH
3
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Ex. 18
Hydrocarbon groups – CH3, –C
3H
7Hydrocarbon groups
– CH3, – C
2H
5, – C
2H
5
Ex. 19
Hydrocarbon groups – C2H
5, – CH
3Hydrocarbon groups – CH
3, – C
2H
5
Ex. 20
O
CH – C – NH – CH3 3
N–MethylethanamideHydrocarbon groups – CH , – CH3 3
O
H – C – NH – CH CH2 3
N–EthylmethanamideHydrogen groups – H, – CH CH2 3
Ex. 21
Groups – CH3, – CH
3Groups – H, – CH
2– CH
3
Ex.22 Compound
)(
CHNHCCHCH||
O
323
I
and
)(
CHCHNHCCH||
O
323
II
are
(A) Functional isomers (B) Position isomers
(C) Metamers (D) All of these
Ans. C
Ex.23 Metamer of the compound P is
P = C – O
O
(A) (B) O – C
O
(C) O – C
O
(D) All of these
Ans. C
2.6 TautomerismTwo compounds have different functional groups due to shiffting of H-atom. The two isomers remain indynamic equilibrium. These two can be isolated.Tautomerism is the phenomenon in which two structural isomers differ in the relative positions of their atomsand are spontaneously interconvertable and can exist in dynamic equilibrium. The two forms in tautomericequilibrium are called tautomers of each other. The interconvertibility of tautomers is a chemical reactionwhich involves making and breaking of bonds.
"manishkumarphysics.in" 45
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e.g.
)(
CHCCH||
O
33
I
)(
CHCCH|
OH
32
II
Tautomers (I) and (II) are structural isomers that are formaly related only by the shift of a hydrogen atom andone or more pi bonds. Tautomerism, thus in a single way, can also be defined as follows type of isomerismoccuring by the migration of atom (generally acidic hydrogen from carbon to electronegative atom andvice-versa) and the movement of a double bond is called tautomerism. Tautomers are true isomers and eitherof the individual tautomeric forms may be isolated.
Structural Requirement for tautomerism
(i) Compound should have electron withdrawing groups such as – CO, – NO2and – C N.
(ii) Compound should have atleast one acidic hydrogen on –carbon of the molecule.
e.g.33 HCCHC
||O
, CH – CH – N3 2
O
O-
+
3
3
CH|
NCHCCH
(iii) If compound has active methylene group than tautomerism is shown by hydrogen of this group
e.g.323 CHCHCCCH
||||OO
Types of tautomerism
(1) Diad SystemIn this system the hydrogen (or the group) migrates from atom number – 1 (i.e. Carbon atom having acidichydrogen to electronegative atom – 2 (i.e., oxygen or nitrogen bonded with multiple bond.)
e.g. (i)21NCH
HNC
(ii) NCR RNC
Alkyl cyanide
(2) Triad SystemIn this system the acidic hydrogen migrates from atom number – 1 to atom number – 3
e.g.33 CHCCH
||O
32 CHCCH
|OH
99.99 % 0.01 %
O OH
99.98 % 0.02 %
HCCH||
O
3 HCCH|
OH
2
99.90 % 0.10 %
The reason is that the C = O double bond of a carbonyl group is a stronger bond than the C = C of the enol.
However, some enols are more stable than the corresponding carbonyl compounds.
OH
99.99 %
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The enol form of phenol is more stable because phenol is aromatic. The enols of –dicarbonyl compounds
are also relatively stable
323 CHCCHCCH||||OO
33 CHCCHCCH
|||OOH
76 %
There are two reasons for the stability of these enols. The first is that they are conjugated but their parent
compounds are not.
The second stabilising effect is the intramolecular hydrogen bond present in enols. Thus enol form is more
stable than keto form by 22 Kcal/mole in 1,3 – diketones.
Ex.24 The tautomer of the following compound P is
P = NH
(A) NH2 (B) NH2 (C) NH2 (D) Both B and C
Ans. D
Ex.25 The correct structure of tautomer of the following compound C
O
CH3
O
is
(A) C
O
CH3
OH
(B) C
OH
CH3
O
(C) C
OH
CH3
O
(D) Both A and B
Ans. D
3. Stereoisomerism(I) The particular kind of isomers that are different from each other only in the way the atoms are oriented in
space are called stereoisomers.
(II) These isomers have same connectivity of atoms and groups.
(III) Stereoisomers have remarkably different physical, chemical and biological properties. e.g. The two
stereoisomers of butenedioic acid are maleic acid and fumaric acid. Fumaric acid is an essential metabolic
intermediate in both plants and animals, but maleic acid is toxic and irritating to tissues.
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Classification of Stereoisomers :
3.1 Configurational Isomers :
(I) These isomers differ in the configuration (The spatial arrangement of atoms that characterises a particular
stereoisomer is called its configuration).
(II) Configurational isomerism arises due to noninterconvertibility at room temperature. Since they are non
interconvertible they can be separated by physical or chemical methods.
3.1.1 Geometrical isomerism :
Isomers which possess the same molecular and structural formula but differ in the arrangement of atoms
or groups in space due to restricted rotation are known as geometrical isomers and the phenomenon is
known as geometrical isomerism.
3.1.2 Conditions of geometrical isomerism :
(I) Geometrical isomerism arises due to the presence of a double bond or a ring structure
(i.e. , – N = N – or ring structure)
Due to the rigidity of double bond or the ring structure to rotate at the room temperature the molecules exist
in two or more orientations. This rigidity to rotation is described as restricted rotation / hindered rotation /
no rotation.
e.g.
(a) (Restricted Rotation)
(b) (Restricted Rotation)
(II) Different groups should be attached at each doubly bonded atom. For example
and are identical but not geometrical isomers.
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On the other hand, following types of compounds can exist as geometrical isomers :
C = C
a
b
a
b, C = C
a
b
a
dor C = C
a
b
d
e
Examples of Geometrical isomers :
(I) Along C = C bond
C = CCH
H
3Br
Hand C = C
CH
H
3H
Br
CH3
C
CH3
H
and
CH3
C
CH3
H
(II) Along C = N – bond
C = N
CH3
H
..
OHand C = N
CH3
H..
OH
C = N
CH3
C H2 5 OH
..
and C = N
CH3
..
OH
C H2 5
N
..
OH
CH3
and N..
OHCH3
N
..
NH2
CH3
C
C H2 5
and N..
NH2CH3
C
C H2 5
(III) Along – N = N – bond
N
....
N
CH3CH3
andN
..
..
N
CH3
CH3
N = N
.. ..
PhPhand N = N
Ph..
.. Ph
(IV) Along bond of cycloalkane
CH3CH3
H H
and
CH3
CH3H
H
CH3CH3
H H
and
CH3
CH3H
H
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CHEMISTRY
CH3
HH3C
H
and
CH3
CH3
HH
H
MeMe
H
and
H Me
Me H
(V) Along C = C in ring structures :
Usually in cycloalkenes double bond has cis configuration. Their trans isomers do not exist due to largeangle strain. But if the ring is large enough a trans stereoisomer is also possible. The smallest transcycloalkene that is stable enough to be isolated & stored is trans-cyclooctene.
3.1.3 Configurational nomenclature in geometrical isomerism
Configuration Criteria Remarks
cis / trans Similarity of groupsIf the two similar groups are on same side of restrictedbond the configuration is cis otherwise trans.
E/Z Seniority of groups
If the two senior groups are on same side of restricted
bond the configuration is Z (Z = zusammen = together)otherwise E (E = entgegen = opposite).
3.1.4 Sequence rules : (cahn - ingold - prelog sequence rules) (CIP)For deciding the seniority of groups following rules are applied :
Rule I : The group with the first atom having higher atomic number is senior. According to this rule the
seniority of atoms is :
I > Br > Cl > S > F > O > N > C > H
Rule II : The higher mass isotope is senior.
Thus (a) – T > – D > – H. (b) – C14H3
> – C12 H3
Rule III : If the first atom of group is identical then second atom is observed for seniority.
e.g.(a) – CH
2Cl > – CH
2OH > – CH
2NH
2> – CH
2CH
3> – CH
3
(b)
OH|
ClC|F
>
2NH|
ClC|F
Rule IV : Groups containing double or triple bonds are assigned seniority as if both atoms were duplicatedor triplicated that
> C = Y as if it were
C – Y| |
(Y) (C)& – C Y as if it were C – Y
|
|
|
|
(Y)
(Y)
(C)
(C)
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e.g. for deciding seniority among – C CH, – CH = CH2, their hypothetical equivalents are
compared.
)HCCfor(CC||
HCC||CC
>
)CHCHfor( 2
HH||
HCC||CC
Rule V : Bond pair is senior to lone pair.Examples :
Ex.28 In which compound, Cis-trans nomenclature cannot be used(A) CH
3– CH = CH – CH
3(B) CH
3– CH = CH – COOH
(C) C = C
CH3
C H2 5Br
Cl
(D) C6H
5– CH = CH – CHO
Ans. C
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Ex.29 Which among the following will show geometrical isomerism
(A)
CH3
DH
CH3
(B)H
H CH3
OH(C) C
6H
5– CH = N – OH (D) all of these
Ans. DEx.30 Which of the following structures will display geometrical isomerism ?
(A) CH3CH = CCl
2(B) CH
3CCl = CBr CH
3
(C) CH3CH = CHBr (D) All of these
Ans. (B) and (C)
(A)
C
C
ClCl
CH3 H
or
C
C
ClCl
CH3H
Both are identical
(B)
C
C
Cl
H C3
CH3
Br
or
C
C
Cl
H C3
CH3
Br
Non – identical
(C)
C
C
CH3
H
H
Br
or
C
C
CH3
H
H
Br
Non – identical
Ex.31 Indicate the configurations of following geometrical isomers
(A) (B)
Ans. (A) E-3-Methylpent-2-ene (B) Cis-hex-2-ene
Ex.32 Indicate the configurations of following geometrical isomers
(A) (B)
Ans. (A) Trans-hex–2–ene (B) Z–3–Methylpent–2–ene
Ex.33 Identify E and Z form of stilbene
(A) (B)
Ans. (A) E (B) Z
3.1.5 Number of Geometrical isomers :Number of geometrical isomers can be found by calculating the number of stereocentres in the compound.(stereocentre is defined as an atom or bond bearing groups of such nature that an interchange of any twogroup will produce a stereoisomer).
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3.1.6 Physical Properties of Geometrical Isomers :
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3.2 Optical Isomers :
3.2.1 Introduction and Definitions :(I) Optical activity & plane-polarised light : Ordinary light is an electromagnetic wave, which has oscilla-
tion in all the directions perpendicular to the path of propagation. When ordinary light is passed through Nicolprism it has all its oscillations in the same plane and is called plane-polarised light.
Certain compounds rotate the plane of polarised light in a characteristic way when it is passed through theirsolutions. These compounds are referred to as optically active compounds. The angle of rotation can bemeasured by an instrument called polarimeter.
(II) dextrorotatory compounds : If the substance rotates plane-polarised light to the right i.e. in clockwise
direction it is called dextrorotatory & indicated by 'd' or (+).
(III) laevorotatory compounds : If light is rotated towards left i.e. in anticlockwise direction the substance
is said to be laevorotatory and indicated by 'l ' or (–).
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(IV) Specific rotation []. Specific rotation is the number of degrees of rotation observed if a
1-dm (10-cm) tube is used and the compound has concentration 1 gm/mL. Thus specific rotation [] is
t][ = C
[] = Specific rotation = observed angle of rotation (degree) = Pathlength (dm)c = concentration (gm/ml) = wavelength (nm)t = temperature (25°C)
3.2.2 Cause of optical activity :
The foundation of modern theory of stereochemistry was laid by Louis Pasteur when he observed two differ-ent kind of crystals, which were mirror images of each other. In aqueous solutions of both types of crystalsshowed optical rotation that was equal in magnitude but opposite in direction. Pasteur believed that thisdifference in optical activity was associated with the three dimensional arrangement of atoms in the twotypes of crystals.Later van't Hoff and LeBel proposed that all the four valencies of carbon are directed towards the four cornersof regular tetrahedron, and if all the four substituent attached to such a carbon are different the resultingmolecule lack symmetry and such a molecule is referred to as asymmetric molecule and asymmetry of themolecule is responsible for optical activity in such organic compounds.
3.2.3 Element of symmetry and concept of molecular dissymmetry/asymmetry andchirality. (I) Plane of symmetry :
It is an imaginary plane which bisects the molecule in two equal halves in such a way that each half of themolecule is the mirror image of the other half.
(II) Centre of symmetry :
The centre of symmetry is defined as the point in a molecule through which if a straight line is drawn from anypart of the molecule this line encounters identical groups at equal distances in opposite direction.
(III) Asymmetric and dissymmetric compounds :
A molecule which does not possess any element of symmetry (there are all 23 elements of symmetry) iscalled asymmetric. A molecule which does not possess plane of symmetry and centre of symmetry is calleddissymmetric.
(IV) Condition for optical activity :
The minimum condition for a compound to show optical activity is molecular dissymmetry i.e. absence ofplane of symmetry and centre of symmetry.
(V) Chirality :
A compound which is non-superimposable to its mirror image is called chiral while a compound which issuperimposable to its mirror image is called achiral.All dissymmetric (or asymmetric) compounds are chiral& vice-versa.
(VI) Chiral centre :
A compound in which a carbon is atttached with four different groups lacks symmetry and is called chiralcarbon or asymmetric carbon. It is represented by C*. e.g.
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(a)
Cl|
CHCHCH 33 (2-chloropropane) :
I has no chiral centre since two groups (a & b) are identical. Its is superimposable on its mirror image II (
III).
(b)
Cl|
HCHCCH 523*
(2-chlorobutane) :
I has one chiral centre it is asymmetric & it is not superimposable to its mirror image II ( III).The necessary condition for chirality is not just the presence of asymmetric carbon atoms but the asymme-try of the molecule as a whole.
Ex.35 Which of the following compounds has chiral carbon ?(A) CH
3– CH = CH – CH
3(B) CH
3– CHOH – CH
2– CH
3
(C) C = N – OHCH3
H(D) All of these
Sol. The starred carbon is chiral carbon because all the four substituents are different CH3– HOH*C – CH
2– CH
3
Ex.36 Mark the asymmetric carbon atoms in the following compounds(A) CH
3– (CHOH)
2– COOH (B) HOCH
2– (CHOH)
4– CHO
Ans. (A)
OHOH
||
COOHCCCH
||
HH
**3 (B)
OHOHOHOHH
|||||
OCCCCCCHO
||||||
HHHHHHH
*****
3.2.4 Projection formula in optical isomerism :(I) Wedge-dash projection formula :
It is a convenient way of depicting three dimensional structure in two dimension. In this projection four bondsof a tetrahedral molecule is shown by two lines (in the plane) one wedge (up the plane) one dash (down theplane)
(II) Fischer projection formula :
It is also a convenient way of depicting three dimensional structure in two dimension.Rules for writing Fischer projection formula :(i) The molecule is drawn in the form of cross (+) with the chiral carbon at the intersection of horizontal &
vertical lines.
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(ii) On vertical line, main chain is taken with first carbon at the top.
(iii) The horizontal lines represent the bonds directed towards the viewer and vertical lines away from theviewer e.g.
(a)OHOH||
CHOHCCH*
2
(glyceraldehyde)
can be represented in two different Fischer projection as
(b)
2
3
NH|
COOHHCCH*
(Alanine)
can be represented in two different Fisher projections as
Ex.37 Draw the wedge–dash projection formula of
(A)Cl H
CF3
CH3
(B)Me OH
NHMe
Ph
Ans. C
CH3Cl
CF3
H (B) C
PhMe
NHMe
OH
3.2.5 Configurational nomenclature in optical isomers :Relative configuration : the experimentally determined relationship between the configurations of twomolecules, even though we may not know the absolute configuration of either. Relative configuration isexpressed by D-L system.Absolute configuration : The detailed stereochemical picture of a molecule, including how the atoms arearranged in space. Alternatively the (R) or (S) configuration at each chirality centre.(I) D - L System (Relative configuration) :This method is used to relate the configuration of sugars and amino acids by the help of enantiomers ofglyceraldehyde. The configuration of (+)-glyceraldenyde has been assigned as D and the compounds withthe same relative configuration are also assigned as D, & those with (-) glyceraldehyde are assigned as L.
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Examples :
Sugars have several asymmetric carbons. Asugar whose highest numbered chiral centre (the penultimasecarbon) has the same configuration as D-(+)-glyceraldehyde (– OH group on right side) is designated as a D-sugar, one whose highest numbered chiral centre has the same configuration as L-glyceraldehyde is desig-nated as an L-sugar.
e.g.
(II) R and S configurations in Fischer projection : (absolute configuration)
Rule I : The priorities of groups which are attached with the asymmetric C-atom are assigned by sequencerule.
Rule II : The lowest priority group is brought to the bottom of Fischer projection by two or even simultaneousexchanges.
Rule III : Then an arrow is drawn from first priority group to second priority group to third priority group. If thearrow is clockwise the configuration assigned to the projection is R & If it is anticlockwise the configurationassigned is S.
Ex.38 (i) (ii)
(iii)
(iv)
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Ex.39 Assign Ror S configuration to each of the following compounds
(A) H OH
CHO
CH OH2
C (B) Br COOH
Cl
H
C (C) H C5 2 H
OH
CH3
C
Ans. (A) R (B) R (C) R
3.2.6 Converting a wedge-dash formula into Fischer projection formula :Step 1 : Decide the priority of groups by sequence rule.Step 2 : Bring the lowest prior group to dash by even simultaneous exchanges.Step 3 : Draw an arrow from first prior group to second prior group till third prior group.Step 4 : If the direction of arrow is clockwise the configuration is R and if anticlockwise it is S.Step 5 : Draw the Fischer projection formula having equivalent configuration to the wedge-dash formula.
Examples : (a)
Here the lowest prior group is already on dash, there is no need for exchanges.
(b)
3.2.7 Enantiomers :Stereoisomers which are non-superimposable mirror images of each other are called enantiomers.e.g. in all the following molecules there is no centre of symmetry or plane of symmetry. Therefore their twomirror image isomers are non-superimposable and hence enantiomers.
Ex.40 2-Chlorobutane :
Ex.41 Chloroiodomethanesulfonic acid :
Enantiomers
I I
SO H3SO H3
H
Ex.42 Lactic acid :
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Ex.43 Alanine :
Ex.44
3.2.8 Properties of enantiomers :
Properties Remarks
(1) Molecular formula Same
(2) Structural formula Same
(3)Stereochemical formula(str. formula with orientation)
Different
(4) Dipole moment Same
(5)Physical properties(m.p., b.p., density, solubility,refractive index etc.)
Same
Chemical properties
(a) with optically inactive compound Same
(b) with optically active compound Different
(6)
3.2.9 Racemic mixture : A mixture of equal amounts of enantiomers is called a racemic mixture or racemic
modification. A racemic modification is always optically inactive when enantiomers are mixed together, therotation caused by a molecule of one enantiomer is exactly cancelled by an equal and opposite rotationcaused by a molecule of its enantiomer.The prefix (±) is used to specify the racemic nature of the particular sample. e.g. (±) Lactic acid, (±) Alanine.
3.2.10 Optical purity : Sometimes we deal with mixtures that are neither optically pure nor racemic mixture. In
these cases we specify the optical purity of the mixture. It is defined as the ratio of its rotation to the rotationof pure enantiomer.
Optical purity =enantiomerpureofrotationoptical
rotationopticalobserved× 100
e.g. If we have some 2-butanol with observed rotation of + 9.72, we compare this rotation with + 13.5rotation of the pure (+) enantiomer.
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optical purity = 5.13
72.9× 100 = 72%.
That means 72% is pure (+) 2-Butanol and 28% is (± mixture)
Total (+) isomer = 72 + 14 = 86%, (–) isomer = 14%
3.2.11 Enantiomeric excess : To compute the enantiomeric excess of a mixture we calculate the excess of
predominant enantiomer as a percentage of the entire mixture. The calculation of enantiomeric excess gives
the same result as the calculation of optical purity.
optical purity = enantiomeric excess
=
d
|d|× 100 = mixtureentire
otheroverenantiomeroneofexcess× 100
Thus for above example
optical purity = enantiomeric excess = d – = 72 %
& d + = 100 %
so 2d = 172 d = 86% & = 14 % (composition of mixture)
3.2.12 Optical diastereomers :The optical isomers which are neither mirror image nor superimposable to each other are called diastere-
omers. Diastereomers have different physical and chemical properties and they can be easily separated by
physical methods.
e.g Let us consider the stereoisomers of 3-chlorobutan-2-ol
There are 4 stereoisomers of 3-chlorobutan-2-ol. In which (I & II) & (III & IV) are enantiomeric pairs. (I & III)
or (I & IV) or (II & III) or (II & IV) all the isomers in each pair are neither mirror image nor superimposable
to each other. Therefore these pairs are optical diastereomers.
3.2.13 Meso compound : A meso compound is one whose molecules are superimposable on their mirror
images even though they contain chiral centres. A meso compound is optically inactive since it has a plane
of symmetry. A meso compound is nonresolvable.
e.g. Let us consider the stereoisomers of 2, 3-Butanediol
In all the possible isomers I & II are enantiomers. But III & IV are not enantiomers since they have plane
of symmetry they are superimposable to each other (all symmetrical compounds are superimposable to
their mirror images).
Thus III & IV are identical & meso compounds.
Thus total stereoisomers of 2, 3-butanediol is 3. Two enantiomers and one meso isomer.
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3.2.14 Number of optical isomers (Stereoisomers) :
e.g. Let us draw the total isomers of
n = 3 (odd chiral centres with similar ends.)
Total isomers = 23–1 = 22 = 4
3.2.15 Reaction of chiral molecules with optically active reagent (optical resolution) :
Resolution refers to the method of separating a racemic mixture into its enantiomeric constituents. A race-
mic mixture is allowed to react with another optically pure compound. This changes a racemic mixture into
a mixture of diastereomers which have different melting and boiling point and solubilities. These can be
separated from one another by conventional method of separation of compounds. The separated diastere-
omers is then broken down to give pure enantiomers. Suppose a racemic mixture (±) A is to be separated. It
is reacted with an optically pure compound (+) B. Thus the schematic diagram for resolution will be.
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Example :
+
+
Diastereomers (Separable)
3.2.16 Optical activity without asymmetric carbon :
(I) Case of allene :
(a) Allenes with even bonds : e.g.
the orbital diagram of this structure will be
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Since the groups at the end of allene are in perpendicular plane, it will not show geometrical isomerism. The
molecule lacks centre of symmetry as well as plane of symmetry. Overall the structure has molecular
dissymmetry which is the sufficient condition for optical activity. The molecule will exist in two enantiomeric
forms.
(b) Allenes with odd bonds : e.g.
the orbital diagram of this structure will be -
The groups at the end of allene structure lie in same plane (ZX plane). Therefore it will have a plane of
symmetry (ZX plane). The molecules lacks molecular dissymmetry & it will not show optical activity hence
optical isomerism. But the compound will exist in two geometrical diastereomeric forms.
(II) Case of spiranes : Asimilar case like allenes is observed in spiranes. The spiranes with even rings and
different groups at terminal carbons show optical activity & optical isomerism, while the spiranes with odd
rings shows geometrical isomerism.
(a) spiranes with even rings :
shows optical isomerism.
(b) spiranes with odd rings :
shows geometrical isomerism.
(III) Case of cycloalkylidene :
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(IV) Case of ortho-ortho-tetrasubstituted biphenyls :
becomes non-planar at room temperature in order to have minimum electronic
repulsion among the substituent. In this orientation (phenyl planes perpendicular to each other) the free
rotation of C – C single bond is restricted and molecule shows optical activity due to molecular disymmetry.
e.g.
3.2.17 Compounds having chiral centres other than carbon :
The tetrahedral atom with four different groups attached to it is a stereocentre (chiral centres). Thus when
atom like nitrogen, phosphorous, sulphur, silicon, germanium has four different groups attached with them
they are chiral and show optical activity & hence optical isomerism. e.g.
3.2.18 Asymmetric nitrogen : Amine inversion
Amines with all different groups attached to N-atom have chiral centres (R1R
2R
3) as nitrogen atom. Since
the geometry of the molecule is tetrahedral it has molecular dissymmetry. It will exist as two enantiomers
but the two enantiomers of amines cannot be separated because they rapidly interconvert to each other.
Therefore they always exist as racemic mixture.
e.g. Ethylmethyl amines 52
..
3 HHCNCH .
Interconvertible ethylmethylamine enantiomers (non resolvable)
3.3. Conformational isomerism3.3.1 Free rotation : For defining free rotation let us consider the bonding in ethane.
We find that the -bond joining the carbon atoms is cylinderically symmetrical about the line joining the twocarbon nuclei ; if the energy does not differ much in different arrangements the molecule can rotate about thiscarbon-carbon bond, we describe this freedom to change by saying that there is free rotation about thecarbon-carbon single bond.
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3.3.2 Conformations :Different arrangements of atoms that can be converted into one another by rotation about single bonds arecalled conformations.
3.3.3 Conformational isomers : There are infinite arrangement (conformations) which arise due free rota-
tionaroundcarbon-carbon bond, different conformations corresponding to energy minima are called confor-mational isomers. The conformational isomerism arises due to free rotation along a bond.
3.3.4 Newman projection : For conformational analysis, a special type of structural formula is convenient to
use which is called newman projection formula and another type is a sawhorse formula.
To write newman projection formula we imagine ourselves taking a view from one carbon atom directly along
the selected bond axis to the next atom. The front carbon and its other bonds are represented as and
those of the back carbon as
3.3.5 Dihedral angle :The angle between C – X and C – Y in X – C – C – Y when it is visualised along C – C bond.
3.3.6 Staggered, eclipsed and skew conformations :(I) The staggered conformation of a molecule is that conformation where the dihedral angle between the
bonds at each atom of carbon-carbon bond is 180°.
(II) In then eclipsed conformation the atoms bonded to carbons at each end of carbon-carbon bond are
directly opposite to one another. The dihedral angle between them is 0°.
e.g.
(III) Skew conformation : All conformations other than staggered or eclipsed are skew conformations.
3.3.7 Factors affecting stability of conformations :
(I) Angle strain : Any atom tends to have bond angles that match those of its bonding orbitals. For example
109°.28 for sp3-hybridized carbon. Any deviation from "normal" bond angles are accompanied by anglestrain.
(II) Torsional strain : Any pair of tetrahedral carbons attached to each other tend to have their bonds
staggered for minimum repulsion between their bonds. Any deviation from the staggered conformation areaccompanied by torsional strain.
(III) van der Waals strain : Non bonded atoms or groups that just touch each other i.e. they are about as
far apart as the sum of their van der Waals raddii, attract each other. If brought any closer together they repeleach other. Such crowding together is accompanied by van der Waals strain.
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3.3.8 Conformational analysis of ethane :Ethane molecule contains a carbon-carbon bond and each carbon is further attached to 3 H-atoms. Itexists in two extreme conformations i.e.
(i) eclipsed conformation
(ii) staggered conformation
The potential energy barrier between the two conformations of ethane is about 12.5 kJ/mol.The potential energy of ethane molecule is at a minimum for the staggered conformation, increase withrotation and reaches a maximum at the eclipsed conformation most ethane molecules,naturally exist in the most stable. staggered conformation. There are only three energy minima, that isethane has only three conformers. Since they are indistinguishable they are degenerate.
Energy profile of Eclipsed and Staggered forms of ethane
3.3.9 Conformational analysis of butane :
If we consider rotation about the C2 - C3 bond of butane, we find that there are six important conformationsshown as I - VI below :
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Energy profile of conforminations of butane :
Conformation
of n-butane
dihedral
angle ()
Torsional
strainvander Waal strain Stability
I : Anti 180° Absent Absent Maximum
II : Partially eclipsed 120° MaximumPresent (between– CH3 & – H grps)
Intermediate - 1
III : Gauche 60° AbsentPresent (betweentwo – CH3 grps)
Intermediate - 2(> Intermediate1)
IV : Fully eclipsed 0° MaximumMaximum (betweentwo – CH3 grps)
Minimum
Strains and stability in n-butane
The stability order will be :Anti > Gauche > Partially eclipsed > Fully eclipsed.
n-Butane exists as three conformational isomers one anti (I) and two gauche (III & V). The gauche
conformers III and V are mirror images of each other and hence are (conformation) enantiomers. Gauche
conformations (III & V) and anti conformation (I) are not mirror images of each other and hence are confor-mational diastereomer n-butane spends the greater part of its time as the anti conformer, and divides thesmaller part equally between the two gauche conformers. As a result of the rapid inter conversion theseisomers can't be separated.
3.3.10 Case of intramolecular hydrogen bonding :In case of G – CH
2– CH
2– OH, where G = – OH, – NH
2, – F, – NR
2, – NO
2, – COOH, – CHO the Gauche form
is more stable than the anti form due to intramolecular hydrogen bonding i.e.
stability : Gauche form > anti form.
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MISCELLANEOUS SOLVED PROBLEMS (MSPS)
1. Write the IUPAC name of following compounds.
(i)
52
23
HOC|
COOH—CH—CH—CH (ii) 3-Bromocyclohexane-1-sulphonic acid
(iii) (iv) 3-Cyano-3-ethoxy-4-nitropentanoyl bromide
Sol. (i) 2-Ethoxybutanoic acid (ii)
(iii) 1,1, 2-Trimethylcyclopentane (iv)
2. Draw the structure of following IUPAC name.
(i) (ii) 3-Methoxycarbonylpropanoic acid
Sol. (i) 3-Ethylpenta-1,4-diyne (ii)
3. Find total number of isomers (including stereo isomers) dimethyl cyclopropane and dimethyl cyclobutane.
(A) 6, 8 (B) 5, 6 (C) 4, 5 (D) 4, 6
Ans. 4, 6
Sol.
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4. has :
(A) plane of symmetry. (B) centre of symmetry.
(C) C3 axis of symmetry. (D) C4 axis of symmetry.
Sol. Centre of symmetry
COS found.
5. How many optically active cycloalkanone are possible with molecular formula C5H8O.
Ans. 6
Sol.
6. and
is / are :
(A) Geometrical isomer (B) Diastereomers
(C) Configurational isomer (D) Enantiomers
Sol. and
are geometrical isomers and they are not mirror image of each other, so they are also known as diastere-
omers and they are interconvertable into each other.
7. For the given compound which of the following statement(s) are correct.
(A) It has four stereoisomer(B) It has 3 stereocentres(C) Only two optically active stereoisomers.(D) On fractional distillation of all stereoisomers only 3 fractions are obtained.
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Sol. Total isomers = (2)3–1 = 4
Optical active isomers = (2)3–1 – 2
13
)2(
= 2
8. The number of stereoisomerss of 1–Bromo–3–methylcyclohexane and 1–Bromo–4–methylcyclohexane are
respectively.
(A) 2,2 (B) 4,0 (C) 4,4 (D) 4,2
Sol. (2 stereocentre then stereoisomers 2n = 22 = 4)
(2 Geometrical stereoisomers)
9. The compound (X) has molecular formula C4H
7Cl. Find out the number of its cyclic isomers (structural and
geometrical only excluding optical isomers).
Ans. 5.
Sol. X = C4H
7Cl U = 5 –
2
8= 1
cis + trans
Total = 5
10. Mark the following as true or false statements.
S1: Eclipsed and staggered forms of ethane can be separated.
S2: The staggered conformations have minimum torsional strain.
S3: Cl–CH
2–CH
2–Cl is a polar compound.
S4: Glycol (HO–CH
2–CH
2–OH) has intramolecular hydrogen bonding in the anti conformation.
(A) F F T T (B) T T T F (C) T T T T (D) F T T F
Sol. S1: Eclipsed and staggered forms of ethane are interconvertable into each other so theycan not be separated.
S4: Glycol (HO–CH
2–CH
2–OH) has intramolecular hydrogen bonding in the gauche conformation.